Solution

The question is a bit tricky.

If you have to choose the NON defective light bulbs, then you must choose among 18 good one.

So, you do have \(\frac{18}{20} * \frac{17}{19} =\frac{153}{190}\)

The correct answer is \(\frac{153}{190}\)

P(neither lightbulb is defective) = P(BOTH lightbulbs work)
= P(1st lightbulb works AND 2nd lightbulb works)
= P(1st lightbulb works) x P(2nd lightbulb works)
= 18/20 x 17/19
= 9/10 x 17/19 [since 18/20 = 9/10]
= 153/190

Answer: 153/190

ASIDE: P(1st lightbulb works) = 18/20, because there are 20 lightbulbs and 18 of them work.
P(2nd lightbulb works) = 17/19 because, once we have selected a working bulb for the 1st selection, there are 19 lightbulbs remaining and 17 of them work.

Cheers,
Brent
_________________

I'm trying to solve this question using the complement method. P(defective) = 1 - P(NOT defective)

P(defective) = 2/20 * 1/19 = 1/190

Inserting in formula :

1/190 = 1 - P(NOT defective)
P(Not defective) = 189/190

What is wrong with this approach?

YOu are going wrong because the answer you are getting is for a different combination..
1-P(both defective) will give you Probability when both are not defective, so it will contain ways when one is defective and one is not AND this i snot the same as the probability
that neither of the lightbulbs selected will be defective?
..
so your complement method should be \(1-P(both -not- defective)-P(just- one -is -defective)=1-\frac{2*1}{20*19}-2*\frac{2*18}{20*19}=1-\frac{1}{190}-\frac{36}{190}=1-\frac{37}{190}=\frac{153}{190}\)..
You are multiplying just one defective by two as it can be defective first and then non-defective or non-defective first and then defective.
Of course here straight method is easier
_________________

18C2 / 20C2
=153/190
_________________

can I have an easy solution?

The solutions above are pretty straightforward. Do you have any questions about a particular step in a particular solution?
_________________

Does simultaneously mean " and " . I mean I use it as an indicator?

"simultaneously" means "at the same time"

This is the author's way of saying the two light bulbs are selected WITHOUT replacement
_________________

how about this?

2 bad "b"
18 good "g"
2 at atime
Hence

bb gg gg gg gg
gg gg gg gg gg

So, 9/10

why can't we use the formula n!/(k!(n-k)!) here?

The combination formula works here also.

P(zero defective bulbs) = (# of ways to select 2 bulbs from the 18 working bulbs)/(# of ways to select 2 bulbs from all 20 bulbs)
= (18C3)/(20C3)
= ((18)(17)/(2)(1))/((20)(19)/(2)(1))
= ((18)(17))/((20)(19))
= 153/190
_________________